Fewpal
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In the fig, the coefficient of friction is 0.2 between the rope and the fixed pulley, and between other surface of contact, m = 0.3. Determine the minimum weight W to prevent the downward motion of the 100N body.

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 From the given fig tanα = 3/4,

cosα = 4/5 & sinα = 3/5, 

Consider equilibrium of block W

RV = 0; R2 = Wcosα ...(1) 

RH = 0; T1 = µR2 + Wsinα ...(2) 

Putting the value of equation(1) in (2)

T1 = µWcos α + Wsinα

= 0.3 X W(4/5) + W(3/5) 

T1 = 0.84W ...(3)

For pulley; T2/T1 = eµ1θ

T2 = T1 X eµ1θ

= 0.84We(0.2 X )

T2 = 1.574W.....(4)

Consider equilibrium of block 100N 

RV = 0; R1 = 100cos α + R2 ...(5) 

R1 = 100cosα + Wcosα = 100(4/5) + W(4/5) 

R1 = 80 + 0.8W ...(6) 

RH = 0; 

T2 = 100sinα –µR1 -µR2 

T2 = 100(3/5) –0.3[(80 + 0.8W) – W(4/5)] 

1.574W = 60 – 24 – 0.24W – 0.24W 

W = 17.53N.

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